Basic concepts of spectroscopy

Almost everything chemists claim to know about a molecule's shape — bond lengths, bond strengths, the identity of a coloured complex — was measured by watching it absorb light. Spectroscopy is the study of how electromagnetic radiation exchanges energy with matter, and a molecule is fussy about which photons it accepts: only those whose energy exactly matches the gap between two of its allowed quantum states get absorbed. Because a molecule stores energy in three very different ways — spinning (rotation), bond-stretching (vibration), and electron rearrangement (electronic transitions) — and because these three energy scales are wildly different in size, different colours of light probe entirely different kinds of motion. This node builds the shared framework — the energy-matching condition, the three spectral regions, and the Beer–Lambert law for measuring how much light is absorbed — that microwave, IR, and UV-Vis spectroscopy each specialise from.

The core insight

A photon is absorbed only when $h\nu$ exactly bridges two quantised molecular energy levels, and because rotational, vibrational, and electronic level spacings differ by orders of magnitude, the microwave, infrared, and UV-Visible regions each probe a distinct kind of molecular motion, with absorption intensity in every case governed by the Beer–Lambert law $A = \varepsilon c l$ .

The core idea

Quantised absorption. A molecule cannot absorb radiation of an arbitrary energy — quantum mechanics restricts it to a discrete ladder of allowed states, and a photon is absorbed only when its energy exactly equals the gap between two of those states.

\Delta E = E_{\text{final}} - E_{\text{initial}} = h\nu = \frac{hc}{\lambda} = hc\tilde{\nu}

where: $\Delta E$ is the energy gap between two allowed molecular states (J), $h$ is Planck's constant, $\nu$ is frequency (Hz), $\lambda$ is wavelength, and $\tilde{\nu} = 1/\lambda$ is the wavenumber (usually in $\text{cm}^{-1}$ ) — the unit chemists prefer for IR and microwave work because it is directly proportional to energy and keeps the numbers manageable.

Three energy scales, three spectral windows. A molecule's total internal energy separates (Born–Oppenheimer approximation) into three roughly independent pieces:

E_{\text{total}} \approx E_{\text{electronic}} + E_{\text{vibrational}} + E_{\text{rotational}}, \qquad E_{\text{electronic}} \gg E_{\text{vibrational}} \gg E_{\text{rotational}}

Because the spacings differ by orders of magnitude, each region of the electromagnetic spectrum wakes up only one kind of motion:

Microwave ( $\sim 1$ – $100~\text{cm}^{-1}$ ): too feeble to touch vibrations or electrons; it only flips a molecule between adjacent rotational levels, and only if the molecule has a permanent dipole moment.
Infrared ( $\sim 400$ – $4000~\text{cm}^{-1}$ ): matches the spacing between vibrational levels of a bond, provided the vibration changes the dipole moment.
UV-Visible ( $\sim 10^4$ – $10^6~\text{cm}^{-1}$ , roughly $200$ – $800~\text{nm}$ ): energetic enough to promote electrons between molecular orbitals.

Measuring how much is absorbed. Regardless of which region is in play, the intensity of absorption obeys the Beer–Lambert law:

A = \log_{10}\left(\frac{I_0}{I}\right) = \varepsilon\, c\, l

where: $A$ is absorbance (dimensionless), $I_0$ and $I$ are the incident and transmitted light intensities, $\varepsilon$ is the molar absorptivity ( $\text{L mol}^{-1}\text{cm}^{-1}$ , how strongly allowed the transition is, and a function of wavelength), $c$ is molar concentration, and $l$ is the path length of the sample cell in cm. The law is universal across microwave, IR, and UV-Vis work; only the physical origin of $\varepsilon$ changes with the region.

Figure

A three-tier ladder diagram showing electronic levels far apart, vibrational sublevels closer together within each electronic level, and rotational sublevels closer still within each vibrational level, with three upward arrows labelled microwave infrared and ultraviolet visible of increasing length

Derivation AFrom transmitted light to molar absorptivity

The goal is to turn a raw laboratory reading — the fraction of light a sample transmits — into the molar absorptivity $\varepsilon$ , the number chemists actually quote to characterise a transition, starting only from the definition of absorbance and the Beer–Lambert law, so that any absorbance-to-concentration or absorbance-to- $\varepsilon$ problem in this unit reduces to the same three lines of algebra.

Step 1. Start from what the instrument measures: the incident intensity $I_0$ entering the cuvette and the transmitted intensity $I$ leaving it. Define the transmittance $T = I/I_0$ , the fraction of light that got through, and define absorbance as $A = \log_{10}(I_0/I) = -\log_{10} T$ . A fully transparent sample has $T = 1$ and $A = 0$ ; a sample that blocks $90\%$ of the light has $T = 0.10$ and $A = 1$ .

Step 2. State the Beer–Lambert law, which says absorbance grows linearly with both how much absorbing substance the light passes through and how strongly that substance absorbs at the chosen wavelength: $A = \varepsilon\, c\, l$ , where $c$ is molar concentration and $l$ is the path length of the cell.

Step 3. Solve for the quantity of chemical interest, the molar absorptivity, by dividing both sides by $c l$ : $\varepsilon = \dfrac{A}{c\,l}$ . Because $A$ is dimensionless, $\varepsilon$ carries units of $\text{L mol}^{-1}\text{cm}^{-1}$ , and it is this number — not $A$ itself, which depends on the sample's concentration and cell — that is a genuine, concentration-independent property of the molecule at that wavelength.

Step 4. Read the physics into the algebra. $\varepsilon$ is large ( $\sim 10^3$ – $10^5$ ) for fully "allowed" transitions such as strongly coloured electronic transitions, and small for "forbidden" ones — so measuring $\varepsilon$ is a direct experimental probe of how allowed a transition is, on top of simply telling you the concentration of an unknown solution once $\varepsilon$ is known.

Worked example

Problem. A photon of wavelength $\lambda = 15~\mu\text{m}$ is absorbed by a diatomic gas. Find (a) its energy in joules and its wavenumber in $\text{cm}^{-1}$ , and (b) which type of molecular transition — rotational, vibrational, or electronic — it most likely excites.

Step 1. Convert wavelength to wavenumber: $\tilde{\nu} = 1/\lambda = 1/(15\times10^{-4}~\text{cm}) = 667~\text{cm}^{-1}$ .

Step 2. Convert wavenumber to energy using $E = hc\tilde{\nu}$ : $E = (6.626\times10^{-34}~\text{J s})(3.0\times10^{10}~\text{cm s}^{-1})(667~\text{cm}^{-1}) = 1.326\times10^{-20}~\text{J}$ .

Step 3. Locate $667~\text{cm}^{-1}$ on the three-window scale. It sits squarely inside the mid-infrared range ( $400$ – $4000~\text{cm}^{-1}$ ) — far too energetic for a pure rotational jump ( $1$ – $100~\text{cm}^{-1}$ ) and far too weak for an electronic one ( $10^4$ – $10^6~\text{cm}^{-1}$ ). This photon excites a vibrational transition — in fact $667~\text{cm}^{-1}$ is close to the real bending-mode frequency of $\text{CO}_2$ .

Competitive stretch

Problem. A $2.5\times10^{-4}~\text{mol L}^{-1}$ solution of a coloured complex sits in a $1.00~\text{cm}$ cuvette, and at $\lambda_{max}$ it absorbs $85\%$ of the incident light. Find $\varepsilon$ at this wavelength.

Step 1. $85\%$ absorbed means $15\%$ transmitted, so $I/I_0 = 0.15$ , giving $A = \log_{10}(1/0.15) = \log_{10}(6.667) = 0.824$ .

Step 2. Apply the Step 3 result of the derivation above: $\varepsilon = A/(cl) = 0.824 / \big[(2.5\times10^{-4}~\text{mol L}^{-1})(1.00~\text{cm})\big] = 3.3\times10^{3}~\text{L mol}^{-1}\text{cm}^{-1}$ . This magnitude ( $\sim10^3$ – $10^4$ ) is typical of a fully allowed electronic transition, matching a strongly coloured complex.

Common traps

Mixing up wavelength and wavenumber trends: $\tilde{\nu} = 1/\lambda$ is directly proportional to energy, so a higher-wavenumber band is higher-energy, while a higher-wavelength band is lower-energy — the reverse of what wavelength intuition suggests.
Treating $E_{\text{rot}}$ , $E_{\text{vib}}$ , and $E_{\text{el}}$ as rigorously independent. They are only approximately additive (Born–Oppenheimer); a real electronic or vibrational transition usually shifts the rotational state too, though that fine structure is outside this course's scope.
Forgetting that $\varepsilon$ is a function of wavelength, $\varepsilon(\lambda)$ , not a single fixed number for a compound — always state the wavelength at which $\varepsilon$ was measured or calculated.
Assuming Beer–Lambert holds unconditionally. It is an ideal law for dilute solutions of one absorbing species; high concentration, aggregation, and stray instrument light all cause real deviations from it.

Ask the tutor: if you want more wavenumber-to-energy conversions, a deeper look at why Born–Oppenheimer lets rotational, vibrational, and electronic energies be treated separately, or extra Beer–Lambert practice, just ask.

Why does chemistry use wavenumber instead of just frequency or wavelength?

Wavenumber $\tilde{\nu} = 1/\lambda$ is directly proportional to energy through $E = hc\tilde{\nu}$ , so a bigger wavenumber always means a bigger energy gap — a convenient, intuitive scale. It also avoids the awkwardly large numbers frequency in hertz produces (frequencies are around $10^{13}$ – $10^{14}$ Hz for IR light) and the awkwardly small numbers wavelength in metres produces. In $\text{cm}^{-1}$ , IR bands land in a friendly $400$ – $4000$ range, which is why every IR spectrum you will see is labelled in wavenumber.

Why does a molecule need a permanent or changing dipole moment to absorb microwave or infrared radiation?

Electromagnetic radiation interacts with matter through its oscillating electric field, and that field can only do work on a molecule if the molecule itself has a fluctuating electric dipole for the field to push and pull on. A rotating polar molecule sweeps its dipole around in a circle, which the field can couple to — hence microwave (pure rotational) activity requires a permanent dipole. A vibrating bond is IR-active only if the vibration itself changes the dipole moment as the bond stretches and compresses; a symmetric stretch that leaves the dipole moment at zero throughout (as in the symmetric stretch of $\text{CO}_2$ ) has nothing for the field to grab, so it is IR-inactive even though it is a real vibration.

Is molar absorptivity epsilon the same thing as absorbance A?

No, and confusing the two is a very common mistake. Absorbance $A$ depends on the specific sample you put in the cuvette — its concentration $c$ and the cell's path length $l$ — so two different solutions of the same compound at different concentrations give different $A$ values. Molar absorptivity $\varepsilon$ is what is left once you divide $A$ by $c$ and $l$ ; it is a fixed property of the compound at a given wavelength, independent of how concentrated your particular sample happens to be, which is why $\varepsilon$ (not $A$ ) is the number tabulated in reference data.

Basic concepts of spectroscopy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads